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I am new to logarithms, and I came across this equation..

$$ \ (3)^{4x} - (3)^{(2x + \log_3(12))} +27 = 0 \ $$

I need a way to simply things as this seems very complex

Edit:

I tried $$ \ (3)^{4x} + 27 = (3)^{(2x + \log_3(12))} \ $$

So $$ \ (3)^{4x} + 3^{3} = (3)^{(2x + \log_3(12))} \ $$

Then, $$ \ log ((3)^{4x} + 3^{3}) = (log(3))(2x + \log_3(12)) \ $$

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  • $\begingroup$ Solve for $y=9^x$. The equation written in terms of $y$ is $y^2-3^{\log_3(12)}y+27=0$. Or $y^2-12y+27=0$. You get that $(y-9)(y-3)=0$. From where either $9^x=y=9$ or $9^x=y=3$. So, either $x=1$ or $x=1/2$. $\endgroup$ – Olivia Jul 9 '17 at 13:07
  • $\begingroup$ Hint: use the substitution $x=3^{2x}$ to get a quadratic. Also bear in mind: $a^{x+y}=a^xa^y$ and $a^{\log_a(x)}=x$ ($0<x$). $\endgroup$ – Shuri2060 Jul 9 '17 at 13:07
  • $\begingroup$ The method you're attempting won't get you very far as you've seen as you can't simplify the expression on the LHS easily. $\endgroup$ – Shuri2060 Jul 9 '17 at 13:09
  • $\begingroup$ Thanks, this is what i was wanting $\endgroup$ – Ravi Prakash Jul 9 '17 at 13:10
  • $\begingroup$ Is there any formula for $ \log(m+n) $ ? $\endgroup$ – Ravi Prakash Jul 9 '17 at 13:13
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$$3^{4x}-3^{2x}\cdot3^{\log_312}+3^3=0$$

$3^{\log_312}$ - is a basic logharithmic rule

then substitute

$3^{2x} = t$

and solve square equation that depends on $t$

I think you can finish it yourself :)

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